Answer:
The direction of the force at A and B is perpendicular to the walls of the container.
The direction of the force at C is down.
The direction of the force in D is up
The direction of the force at E is to the left.
The attached figure shows the forces exerted by the water at points A, B, C, D and E.
Explanation:
The water is in contact with the bowl and with the fish. It exercises at points A, B, C, D and E, but the direction is different from the force.
The fish has a buoyant force on the water and that direction is up. The direction of at point D is up.
The column of water on the fish has a downward force, therefore the direction of the force at point C is down. The water column to the right of the fish has a force to the left, and the direction at point E is to the left.
The water will exert a force on the walls of the container and this force at points A and B is a on the walls of the container.
Answer:
Work done on an object is equal to
FDcos(angle).
So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.
However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.
But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?
Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.
Explanation:
Answer:
k1 + k2
Explanation:
Spring 1 has spring constant k1
Spring 2 has spring constant k2
After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.
x1 = x2
Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are
k1 = F1/x -> F1 =k1*x
k2 = F2/x -> F2 =k2*x
While F = F1 + F2
Substitute equation of F1 and F2 into the equation of sum of forces
F = F1 + F2
F = k1*x + k2*x
= x(k1 + k2)
Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)
Considering the general equation of spring forces (Hooke's Law) F = kx,
The effective spring constant for the system is k1 + k2
